using fermat's little theorem show that each prime divisor of (2^p) -1, where p is a prime is greater than p.

any help on this would be appreciated. thanks in advance!

Since it means . Let be the order of modulo then , but by properties of orders we know since it means thus the is the order of modulo . But by Fermat's little theorem we know and hence by properties of order. In particular, this means . Thus, , a prime divisor of , is greater than .

thanks for the response but I don't quite understand still. Is there a way you could explain it some more or provide a link to a place where I might find something to help me understand it?
Thanks

Given a prime and an integer (with ) we define the order of mod to be the least positive exponent such that . Of course such a can always be found and furthermore because and is the least (smallest) such exponent. For example, take and then because . Now there is a simple theorem with orders. Suppose is a positive integer such that then *. This theorem is telling us that the order must divide any exponent which raises congruent to . So in the above proof we concluded that but since is a prime it forces .

*)In case you want to see the proof it is simple. We can write where by the division algorithm. That means but so but that is impossible since and is least so . Which means . And so clearly .